Continuous Logic and Probability Algebras THESIS

Transcription

1 Continuous Logic and Probability Algebras THESIS Presented in Partial Fulfillment of the Requirements for the Degree Master of Science in the Graduate School of The Ohio State University By Fan Yang, B.S. Graduate Program in Mathematics The Ohio State University 2016 Master's Examination Committee: Christopher Miller, Advisor Timothy Carlson

2 Copyright by Fan Yang 2016

3 Abstract Continuous logic is a multi-valued logic where the set of truth values is the unit interval [0, 1]. It was developed recently as a framework for metric structures, which consist of complete, bounded metric spaces on which there are distinguished elements and uniformly continuous functions. There are many parallels between continuous logic and first-order logic: 0 corresponds to true and 1 to false, sup and inf take the place of quantifiers, and uniformly continuous functions on [0, 1] replace connectives. Instead of a distinguished equality symbol, there is a distinguished predicate for the metric. Familiar theorems of first-order logic, such as completeness, compactness, and downward Löwenheim-Skolem, have modified counterparts in continuous logic. We present these results in comparison to those in first order logic and prove that the class of probability algebras (probability spaces modulo null sets where the distance between two events is their symmetric difference) is axiomatizable in continuous logic. ii

4 Vita Bozeman High School B.S. Mathematics, Carnegie Mellon University Graduate Teaching Associate, Department of Mathematics, The Ohio State University Fields of Study Major Field: Mathematics iii

5 Table of Contents Abstract... ii Vita... iii Chapter 1: Introduction... 1 Chapter 2: Preliminaries... 3 Chapter 3: Axioms of continuous logic... 9 Chapter 4: Probability algebras...17 Chapter 5: Continuous logic versus first-order logic...53 Chapter 6: Conclusion...59 References...61 iv

6 1. INTRODUCTION Multi-valued logics are formal systems that deviate from classical logic by allowing for more than two truth values. An early version of multi-valued logic appeared in 1920 with the development of a three-valued logic by Łukasiewicz. This was extended to an infinitevalued propositional logic for which Pavelka proved a completeness result [3]. The next big step for multi-valued logics occurred in 1966, when Chang and Keisler generalized the set of truth values to be any compact Hausdorff space [5]. Among their results were suitable rephrasings of the compactness theorem and the Downward Löwenheim-Skolem theorem. Ultimately, their construction proved to be too general for applications to other areas of mathematics and little was done until the need arose in model theory to find a framework in which to study metric structures. Metric structures consist of complete, bounded metric spaces on which there are distinguished elements and uniformly continuous functions. This paper presents a recent incarnation of multi-valued logic called continuous firstorder logic, or simply continuous logic, introduced by Ben Yaacov and Usvyatsov in [4] to study metric structures. It is a special case of Chang and Keisler s logic in [5] in the sense that the set of truth values is restricted to the unit interval [0, 1]. However, it differs from [5] by considering structures equipped with a distinguished metric instead of the traditional equality relation. In the same way that the equality symbol = in first-order logic is always interpreted to be real equality, the language of continuous logic contains a distinguished binary predicate d that is always interpreted to be the metric. First-order model theory coincides with continuous model theory if we equip the underlying set of any first-order 1

7 model with the discrete metric. Thus continuous logic is an extension of first-order logic in the sense that it allows more structures to be models; see, e.g., Iovino [11]. These structures include Hilbert spaces, Banach spaces, and probability algebras. Probability algebras, which arise from taking quotients of probability spaces modulo their null sets and become complete metric spaces under d(a, b) = µ(a b), where µ is the probability measure, will be the focus of the first part of this paper. Although continuous logic was developed by model theorists for model theoretic uses, some research about its properties as a logic has also been done. Notably, Ben Yaacov and Pederson formulated in [3] a set of axioms that yields a version of the completeness theorem. A comparison of the completeness theorem and other aspects of continuous logic to their counterparts in first-order logic will be presented in the second part of this paper. 2

8 2. PRELIMINARIES Let (M, d) be a complete metric space with diameter 1. Definition. A predicate on M is a uniformly continuous function P : M n [0, 1]. A function on M is a uniformly continuous function f : M n M. Definition. A metric structure M consists of a complete metric space (M, d) with diameter 1, a set {P i : i I} of predicates on M, a set {f j : j J} of functions on M, and a set {c k : k K} of distinguished elements of M. We denote it by M = (M, P i, f j, c k : i I, j J, k K). Convention. We will regard constants as 0-ary functions and suppress explicit mention of them in future definitions. We treat d as a binary predicate since it is a uniformly continuous function from M 2 [0, 1]; this is proven in 4.9. The set of natural numbers {0, 1,...} is denoted by N. Here are some examples of metric structures. (1) Let (M, d) be any complete metric space of diameter 1. Then M = (M, d) is a metric structure with no functions and whose sole predicate is the binary predicate d. (2) Let M = (M, P i, f j, c k : i I, j J, k K) be a structure in first-order logic. Equip M with the discrete metric: for all x, y M, d M (x, y) = 1 if and only if x y. This makes M a complete metric space with diameter 1. Predicates take 3

9 values in {0, 1} [0, 1]. Since the metric is discrete, all predicates and functions are trivially uniformly continuous. Under the intended interpretation of 0 as and 1 as, d becomes a generalization of equality: x = y is true in M if and only if d M (x, y) = 0. In this sense, first-order models are a special case of metric structures. (3) Given a probability space (X, B, µ), we can construct a probability structure, a metric structure based on the metric space (M, d) where M consists of elements of B modulo 0 and d is the measure of symmetric difference. We take µ to be a unary predicate and the Boolean operations,, and c to be functions on M. We will study an axiomatization of this class of structures in the fourth chapter of this paper. Definition. Let (M, d), (M, d ) be metric spaces. Let s : M M be a function. δ s : (0, 1] (0, 1] is a modulus of uniform continuity for s if for all ɛ (0, 1] and x, y M, we have d(x, y) < δ s (ɛ) = d (s(x), s(y)) < ɛ. Note that s is uniformly continuous if and only if it has a modulus of uniform continuity. Convention. If (M, d) is a metric space and f : M n M is an n-ary function, then we always equip M n with the maximum metric: given x = (x 1,..., x n ), y = (y 1,..., y n ) M n, we use dmax(x, y) = max{d(x i, y i ) : i = 1,..., n}. If (M, d) is a metric space and P : M n [0, 1] is a n-ary predicate, then we always equip [0, 1] with the usual metric: given a, b [0, 1], d [0,1] (a, b) = a b. Definition. A continuous signature is a quadruple L = (R, F, η, G) such that (1) R contains a distinguished binary predicate d (2) R F = 4

10 (3) η : R F N (4) G = {δ s : (0, 1] (0, 1] : s R F} R is the set of relation symbols, F is the set of function symbols, and (R \ {d}) F is the set of nonlogical symbols of L. The arity of a relation or function symbol s is given by η(s). G is the set of moduli of uniform continuity for each predicate and function symbol. The moduli of uniform continuity are not syntactic objects; rather, they are functions (0, 1] (0, 1] fixed by the signature. The cardinality of L, denoted by L, is the cardinality of the set of nonlogical symbols of L. The logical symbols of any signature L consist of the following: a binary connective,. two unary connectives, and 1, two quantifiers, sup and inf, countably many variables 2 V = {v 0, v 1,...}. Since we always interpret d to be the metric, d is also considered a logical symbol. This parallels the treatment of the equality symbol in first-order logic Definition. Let L = (R, F, η, G) be a continuous signature. An L-structure is an ordered pair M = (M, ρ) where M is a set and ρ is a function on R F such that (1) for each f F, ρ assigns a map f M : M η(f) M (2) for each P R, ρ assigns a map P M : M η(p ) [0, 1] (in particular, ρ assigns d to a complete metric d M : M M [0, 1]) (3) for each f F, δ f is a modulus of uniform continuity for f: for all ɛ (0, 1] and x = (x 1,..., x η(f) ), y = (y 1,..., y η(f) ) M η(f), dmax(x, y) < δ f (ɛ) = d M (f(x), f(y)) < ɛ 5

11 (4) for each P R, δ P is a modulus of uniform continuity for P : for all ɛ (0, 1] and x = (x 1,..., x η(p ) ), y = (y 1,..., y η(p ) ) M η(p ), dmax(x, y) < δ P (ɛ) = P (x) P (y) < ɛ Remark. Part (2) of the above definition establishes the set of truth values as [0,1]. Definition. Fix a continuous signature L = (R, F, η, G). L-terms are defined inductively. Variables and constant symbols are L-terms. If f F is an n-ary function symbol and t 1,..., t n are L-terms, then f(t 1,..., t n ) is an L-term. Atomic L-formulas are expressions of the form P (t 1,..., t n ), where P R is an n-ary predicate symbol and t 1,..., t n are L- terms. The class of L-formulas is the smallest class of expressions that contains all atomic formulas, is closed under,., and 1, and satisfies the following condition: If φ is an 2 L-formula and x is a variable, then sup x φ is an L-formula Definition. Let M = (M, ρ) be an L-structure. An M-assignment is a function σ : V M. Given an M-assignment σ, x V, and a M, define the M-assignment σ a x as follows: for all y V, a σx(y) a := σ(y) if x = y otherwise Fix an M-assignment σ. Define t M,σ, the interpretation under σ of a term t in M, inductively in the same way as in first-order logic. Let φ be a formula. Define M(φ, σ), the value of φ in M under σ, by induction on formulas: (1) M(P (t 1,..., t η(p ), σ) := P M (t M,σ 1,..., t M,σ η(p ) ) (2) M(α. β, σ) := max(m(α, σ) M(β, σ), 0) (3) M( α, σ) := 1 M(α, σ) 6

12 (4) M( 1α, σ) := 1 M(α, σ) 2 2 (5) M(sup x α, σ) := sup{m(α, σ a x) : a M} When φ is a sentence, we write M(φ) instead of M(φ, σ) We introduce the following abbreviations for use in future examples. Let 1 stand for (φ. φ) for any sentence φ. Let ( 1 2 )n stand for }{{ 2} n times This notation is natural because for any formula φ, the truth value of (φ. φ) in any model under any assignment is 1: M( (φ. φ), σ) = 1 M(φ. φ, σ) = 1 max(m(φ, σ) M(φ, σ), 0) = 1. Similarly, for any n, the truth value of ( 1 2 )n is always ( 1 2 )n. We can use this to construct a sentence α whose truth value in any model M is 3 4. First, define a binary connective + 0 by φ + 0 ψ = (ψ φ). Then M(φ + 0 ψ, σ) = min{m(φ, σ) + M(ψ, σ), 1} for all models M and assignments σ. Note that + 0 is classically equivalent to conjunction. Now take α to be ( 1 2 )2. Then M(α, σ) = min{ 1 + 1, 1} = 3. In this way, for every dyadic number k 2 n where k, n N, k 2 n, we can produce a sentence whose truth value is k, so the set of truth values generated by {,,. 1 k } from 1 is D = { : k, n N, k 2 n }. 2 n 2 2 n In propositional logic, a chosen set of connectives must be functionally complete; every Boolean function should be representable in terms of a propositional formula formed from these connectives. The dual notion of functional completeness in continuous logic is fullness. In general, for each n 1, an n-ary connective f is a continuous function from 7

13 [0, 1] n [0, 1]. A system of connectives is a family F = (F n n 1) where each F n is a set of n-ary connectives. F is closed if the following hold: (1) for all n, F n contains the projection (a 1,..., a n ) a j for each j = 1,..., n. (2) for all m, n, if u F n and v 1,..., v n F m, then w : [0, 1] m [0, 1] defined by w(a) = u(v 1 (a),..., v n (a)) is in F m. For each system of connectives F, let F denote the smallest closed system of connectives containing F. F is full if for each n, F is dense in the set C n of all n-ary connectives [0, 1] n [0, 1] with respect to the topology on C n obtained from the metric d(f, g) = sup x [0,1] n f(x) g(x). In other words, for all n 1, ɛ > 0, and f C n, there exists g F such that for all x [0, 1] n, f(x) g(x) < ɛ. Instead of adding a symbol for each connective, which would render the language uncountable, we may choose any full system of connectives, allowing us to approximate any connective to arbitrary precision. Fact. (6.6 in [2]) The system of connectives {,., 1 } is full. 2 Thus we choose {,., 1 }, which is the analogue of {, }, a popular choice of func- 2 tionally complete connectives in first-order logic. Interestingly, although {,, } is functionally complete in first-order logic, it is not full in continuous logic with and as defined in 3.3; by 1.7 in [4], all n-ary connectives constructed using,, are 1-Lipschitz in each argument, but 1 2 is 2-Lipschitz, so 1 2 cannot be constructed from,, Definition. Let M be an L-structure, σ an assignment, and Σ a set of L-formulas. (M, σ) is a model of Σ, written (M, σ) Σ, if for all φ Σ, M(φ, σ) = 0. We will abbreviate (M, σ) {φ} by (M, σ) φ. Σ is satisfiable if it has a model. We write Σ φ if every model of Σ is a model of φ. φ is valid if φ. 8

14 3. AXIOMS OF CONTINUOUS LOGIC Definition. Given a term t, a variable x, and a formula α, the substitution of t for x in α, denoted α[t/x], is the result of replacing every free occurrence of x in α by t. Define φ ψ := ψ. φ. The following axiom system from [3] for continuous logic allows us to prove the completeness result in 5.3. The axioms of continuous logic are the universal closures of: (1) α (β α) (2) (α β) ((β γ) (α γ)) (3) ((α β) β) ((β α) α) (4) ( α β) (β α) (5) ( 1α α) 1α 2 2 (6) 1α ( 1α α) 2 2 (7) sup x (α β) (sup x α sup x β) (8) sup x α α[t/x], if no variable in t is bound by a quantifier in α (9) α sup x α, if x is not free in α (10) d(x, x) (11) d(x, y) d(y, x) (12) d(y, z) (d(x, y) d(x, z)) (13) For each f F, ɛ (0, 1], and r, q D, if r > ɛ and q < δ f (ɛ), then ( d(x i, y i ) q)) (r d(f(x 1,..., x η(f) ), f(y 1,..., y η(f) ))) i=1,...,η(f) 9

15 (14) For each P R, ɛ (0, 1], and r, q D, if r > ɛ and q < δ P (ɛ), then ( d(x i, y i ) q)) (r P (x 1,..., x η(p ) ) P (y 1,..., y η(p ) ) ) i=1,...,η(p ) The only rule of inference in continuous logic is modus ponens α, α β β Formal deductions and the provability relation are defined in the usual way Interpreting implication Let M be an L-structure, σ an assignment, and φ, ψ L-formulas. By definition, M(φ ψ, σ) = M(ψ φ, σ) = max(m(ψ, σ) M(φ, σ), 0), so (M, σ) φ ψ M(ψ, σ) M(φ, σ). In words, φ ψ holds in (M, σ) if and only if ψ is at most as true as φ. If we restrict the set of truth values to {0, 1}, then this says that the only instance where φ ψ does not hold in (M, σ) is when M(ψ, σ) > M(φ, σ), i.e. when M(ψ, σ) = 1 and M(φ, σ) = 0. An interesting phenomenon occurs when ψ is φ. In this case, (M, σ) φ φ 1 M(φ, σ) = M( φ, σ) M(φ, σ) M(φ, σ) 1 2. Thus for a formula φ to imply its own negation in (M, σ), it only needs to be more false than 1 2 in (M, σ) Defining disjunction 10

16 In first-order logic, there are many ways to define disjunction in terms of and. For example, we could take φ 1 ψ := ψ φ and φ 2 ψ := (ψ φ) φ. With the semantics of continuous logic, M(φ 1 ψ, σ) = M( ψ φ, σ) = max(m(ψ, σ) + M(φ, σ) 1, 0) M(φ 2 ψ, σ) = M((ψ φ) φ, σ) = min(m(φ, σ), M(φ, σ)) If we restrict the range of M(, σ) to {0, 1}, then M( 1, σ) M( 2, σ), reflecting the fact that 1 and 2 are equivalent in first-order logic. To see that they are not equivalent in continuous logic, consider φ = 1 2 as defined in 2.3. Since M(φ, σ) = 1 2, M(φ 1 φ, σ) = max(m(φ, σ) + M(φ, σ) 1, 0) = max( , 0) = M(φ 2 φ, σ) = min(m(φ, σ), M(φ, σ)) = min( 1 2, 1 2 ) = 1 2. Our choice of disjunction should give the same truth value to φ and to φ φ. Furthermore, for all models M and assignments σ, it should satisfy (M, σ) φ ψ (M, σ) φ or M ψ and M(φ ψ, σ) = min(m(φ, σ), M(ψ, σ)), so the most natural choice is = 2 (or anything equivalent to it in continuous logic). Similarly, conjunction should satisfy (M, σ) φ ψ (M, σ) φ and M ψ and M(φ ψ, σ) = max(m(φ, σ), M(ψ, σ)), both of which are achieved by taking φ ψ := ( φ ψ). These remarks are summarized below. 11

17 3.3. We introduce the following abbreviations: disjunction φ ψ := (ψ φ) φ conjunction φ ψ := ( φ ψ) logical distance φ ψ := (φ ψ) (ψ φ) inf x φ := sup x φ Under the semantics of continuous logic, (1) M(φ ψ, σ) = max(m(ψ, σ) M(φ, σ), 0) (2) M(φ ψ, σ) = min(m(φ, σ), M(ψ, σ)) (3) M(φ ψ, σ) = max(m(φ, σ), M(ψ, σ)) (4) M( φ ψ, σ) = M(φ, σ) M(ψ, σ) (one-dimensional Euclidean distance) (5) M(inf x φ, σ) = inf{m(φ, σx) a : a M} If we restrict the truth values to {0, 1}, the interpretations of sup, inf,,, and coincide with those of,, disjunction, conjunction, and implication, respectively, in first-order logic. For example, (M, σ) inf x φ M(inf x φ, σ) = 0 there exists a M such that M(φ, σx) a = 0. When it is easier to do so, we will write φ ψ instead of φ ψ. Remark. Under the abbreviations in 3.3, axiom 3 states that is commutative. Axioms 1 4 are the axioms for Łukasiewicz propositional logic [3]. Their relation to the axioms of classical propositional logic is explored in

18 Axioms 5 and 6 establish the behavior of 1 2. Recall from 2.3 that α + 0 β := (β α) and M(α + 0 β, σ) = min(m(α, σ) + M(β, σ), 1) for any L-structure M and M- assignment σ. In particular, M( 1 2 α α, σ) = min(m(1 2 α, σ) + M(1 α, σ), 1) 2 = min( 1 2 M(α, σ) + 1 M(α, σ), 1) 2 = min(m(α, σ), 1). Axioms 7 9 establish the behavior of sup and are direct translations of the quantifier axioms in first-order logic [7]. Axioms are translations of properties of = in first-order logic. Recall the equality axioms of first-order logic (from [1]): (i) reflexivity: x = x (ii) substitutivity: x = y (φ φ ) for any formula φ, where φ is obtained from φ by replacing some free occurrences of x with y. Let M be an L-structure and let σ be an assignment. If we replace a = b by d(a, b), then axiom 10 is a direct translation of (i). Consequences of (ii) include symmetry, (x = y) (y = x), and transitivity, (x = y) ((x = z) (y = z)), which are encoded in axioms 11 and 12, respectively. So the property of = being an equivalence relation translates, via axioms 10 12, into d being a pseudo-metric. The reason why we do not need to specify that d is a genuine complete metric is given in 3.4. Equality axiom (ii) also makes = a congruence relation; for each predicate P, (x = y) (P (x, z) P (y, z)). By 3.1, 13

19 (M, σ) (d(x, y)) (P (x, z) P (y, z)) M(P (x, z) P (y, z), σ) M(d(x, y), σ) max(m(p (y, z), σ) M(P (x, z), σ), 0) d M (x, y). We also have max(m(p (x, z), σ) M(P (y, z), σ), 0) d M (y, x) = d M (x, y), so P M (x, z) P M (y, z) = M(P (x, z), σ) M(P (y, z), σ) d M (x, y). In other words, predicates should be 1-Lipschitz with respect to d, although 1-Lipschitz is relaxed to uniformly continuous. A similar argument using (x = y) (f(x, z) = f(y, z)) translated to d(x, y) (d(f(x, z), c) d(f(y, z), c)) shows that functions should be 1-Lipschitz, which is likewise relaxed to uniformly continuous. To express these requirements, axioms 13 and 14 define the moduli of uniform continuity for each function and predicate symbol We now introduce a more general notion of metric structure and summarize its relation to metric structures. Definition. Using the notation from 2.1, an L-pre-structure is an L-structure such that ρ assigns d to a pseudo-metric. Definition. Let L = (R, F, η, G) be a continuous signature. Let M and N be L-prestructures with underlying sets M and N, respectively. A function h : M N is an elementary L-morphism if it satisfies the following conditions: (1) for all f F and a 1,..., a η(f) M, h(f M (a 1,..., a η(f) )) = f N (h(a 1 ),..., h(a η(f) )). 14

20 (2) for all P R and a 1,..., a η(p ) M, P M (a 1,..., a η(p ) )) = P N (h(a 1 ),..., h(a η(p ) )). (3) for all M-assignments σ and formulas φ, M(φ, σ) = N (φ, h σ). The definitions of assignment, interpretation, and value in L-pre-structures are the same as those in L-structures from 2.2. Given an L-pre-structure M, an assignment σ, and a set of L-formulas Σ, we say that (M, σ) is a pre-model of Σ if for all φ Σ, M(φ, σ) = 0. We say that a set of formulas Σ is pre-satisfiable if it has a pre-model. Theorem. (6.9 from [3]) Let L be a continuous signature and let M be an L-pre-structure. Then there exists an L-structure M and an elementary L-morphism of M into M. M can be constructed by taking the quotient space of M under the equivalence relation x y d(x, y) = 0. Corollary. A set of formulas Γ is satisfiable if and only if Γ is pre-satisfiable. Thus L-pre-structures and L-structures are indistinguishable in continuous logic. The analogue of this fact in first-order logic is that = is a congruence relation, so the quotient of a first-order structure K by = K is indistinguishable from K in first-order logic. We end this section with definitions of elementary equivalence and elementary substructure. They coincide with the definitions from first-order logic when the set of truth values is restricted to {0, 1}. Definition. Let M and N be L-structures. M and N are elementarily equivalent if M(φ) = N (φ) for all L-sentences φ. M is an elementary substructure of N if M N 15

21 and for all formulas φ(x 1,..., x n ) and a 1,..., a n M, M(φ, σ) = N (φ, σ ) where σ is any M-assignment and σ is any N -assignment such that σ(x i ) = σ (x i ) = a i for each i = 1,..., n. Definition. Let L be a continuous signature. A sentence is an L-formula with no free variables. A theory T is a set of sentences. The theory of an L-structure M, denoted by Th(M), is the set of sentences true in M. Remark. If M and N are elementarily equivalent, then Th(M) = Th(N ). The converse also holds: Suppose for L-structures M and N, Th(M) = Th(N ). Let φ be a sentence and let q = M(φ). By 2.3, there exists an L-sentence ψ such that in all L-structures Q, Q(ψ) = q. In particular M(ψ) = q, so M(φ ψ) = 0. Since Th(M) = Th(N ), N (φ ψ) = 0. Since N (ψ) = q, we have N (φ) = N (ψ) = q = M(φ), so M and N are elementarily equivalent. 16

22 4. PROBABILITY ALGEBRAS Definition. [10] A Boolean algebra is a set A that includes two distinguished constants 0 and 1 and on which there are defined two binary operations and, a unary operation c which satisfy the following rules: (1) x x x = x (2) x x x = x (3) x y x y = y x (4) x y x y = y x (5) x y (x y) y = y (6) x y (x y) y = y (7) x y z (x y) z = x (y z) (8) x y z (x y) z = x (y z) (9) x y z x (y z) = (x y) (x z) (10) x y z x (y z) = (x y) (x z) (11) x x x c = 1 (12) x x x c = 0 (13) 0 1 A Boolean algebra becomes partially ordered if we define x y to be x y = x (reflexive by (2), antisymmetric by (4), transitive by (8)). A Boolean algebra A is a 17

23 Boolean σ-algebra if for each sequence x n of elements of A there is a least element x A such that x n x for all n. We denote this least upper bound by sup n x n Facts about Boolean algebras. Let B be a Boolean algebra. For all a, b B, let a \ b := a b c For all a B, a 0 = a, a 0 = 0, a 1 = 1, and a 1 = a disjoint unions. For all a 0 a 1 a 2 B, a 2 \ a 0 = (a 2 \ a 1 ) (a 1 \ a 0 ) and (a 2 \ a 1 ) (a 1 \ a 0 ) = 0. Furthermore, i < j {0, 1, 2} implies a i \ a j = 0. Proof. Observe (a 2 a c 1) (a 1 a c 0) = (a 2 a c 1) ((a 2 a 1 ) a c 0) = (a 2 a c 1) (a 2 (a 1 a c 0)) = a 2 (a c 1 (a 1 a c 0)) = a 2 ((a c 1 a 1 ) (a c 1 a c 0)) = a 2 (a c 1 a c 0) = a 2 (a 1 a 0 ) c = a 2 a c 0 = a 2 \a 0 and (a 2 \a 1 ) (a 1 \a 0 ) = (a 2 a c 1) (a 1 a c 0) = a 2 (a c 1 a 1 ) a c 0) = 0. Now let i < j {0, 1, 2}. Then a i \ a j = a i a c j = (a i a j ) a c j = a i (a j a c j) = a i 0 = b b n implies b b n = b n Proof. Suppose b b n. By definition b b n = b, so by axiom 5 b b n = (b b n ) b n = b n For all a, b, c B, (a b) c = (a c) (b c). Proof. (a b) c = ((a b c ) (b a c )) c = ((a b c ) c) ((b a c ) c) = ((a c) b c ) ((b c) a c ) = [((a c) b c ) 0] [((b c) a c ) 0] = [((a c) b c ) (a 0)] [((b c) a c ) (b 0)] = [((a c) b c ) (a (c c c ))] [((b c) a c ) (b (c c c ))] = [((a c) b c ) ((a c) c c )] [((b c) a c ) (b (c c c ))] = ((a c) (b c c c )) ((a c c c ) (b c)) = ((a c) (b c) c ) ((a c) c (b c)) = (a c) (b c). 18

24 For all a, b, c B, a c (a b) (b c). Proof. Let a, b, c B. Then (a c c ) ((a b) (b c)) = (a c c ) ((a b c ) (a c b) (b c c ) (b c c)) = ((a c c ) (a b c )) ((a c c ) (a c b)) ((a c c ) (b c c )) ((a c c ) (b c c)) = (a c c b c ) 0 (a b c c ) 0 = ((a c c ) b c ) ((a c c ) b) = (a c c ), so a c c (a b) (b c). By symmetry of, a c c (a b) (b c), so a c ((a b) (b c)) = ((a c c ) (a c c)) ((a b) (b c)) = ((a c c ) ((a b) (b c)) ((a c c) ((a b) (b c))) = (a c c ) (a c c) = a c. Hence, a c (a b) (b c) For all a, b B, if a \ b = 0, then a b. Proof. Let a, b B and suppose a\b = 0. Then a = a 1 = a (b b c ) = (a b) (a b c ) = (a b) (a \ b) = (a b) 0 = a b, so a b For all a, b, c B, (a b) c = a (b c) Definition. A probability space is a triple (X, B, µ) where X is a set, B is a σ-algebra of subsets of X, and µ is a σ-additive measure on B such that µ(x) = Proposition. Let (X, B, µ) be a probability space. Define the relation µ on B by A µ B µ(a B) = 0. Then µ is an equivalence relation. Proof. Clearly µ is reflexive and symmetric. Let A, B, C B such that A µ B and B µ C. Then µ(a B) = µ(b C) = 0. Since A \ B and B \ A are disjoint, 0 = µ(a B) = µ((a \ B) (B \ A)) = µ(a \ B) + µ(b \ A), so µ(a \ B) = µ(b \ A) = 0. Similarly, µ(b \ C) = µ(c \ B) = 0. Since A \ C (A \ B) (B \ C) and C \ A (C \ B) (B \ A), µ(a C) = µ((a \ C) (C \ A)) = µ(a \ C) + µ(c \ A) 19

25 µ(a \ B) + µ(b \ C) + µ(c \ B) + µ(b \ A) = 0. Thus A µ C, so µ is transitive, and hence an equivalence relation. Notation. Denote the equivalence class of A B by [A] µ. Let ˆB = {[A] µ : A B} be the set of equivalence classes and define the following operations on ˆB: For all A, B B, let [A] c µ = [A c ] µ, [A] µ [B] µ = [A B] µ, and [A] µ [B] µ = [A B] µ. For all (A n ) n N B N, let [A n ] µ = [ A n ] µ and [A n ] µ = [ A n ] µ. n=0 n=0 n=0 n= Proposition. As defined above, c,, and are well-defined. Proof. Let (X, B, µ) be a probability space and let ˆB = {[A] µ : A B}. Let [A] µ = [B] µ ˆB. Since A B = A c B c, 0 = µ(a B) = µ(a c B c ), so [A c ] µ = [B c ] µ. Next, let (A n ) n N, (B n ) n N B N such that [A n ] µ = [B n ] µ for all n N. Then µ(( A i ) ( B j )) j=0 = µ((( A i ) \ ( B j )) (( B j ) \ ( A i ))) j=0 j=0 = µ( (A i \ ( B j )) (B j \ ( A i ))) j=0 j=0 µ( (A i \ ( B j ))) + µ( (B j \ ( A i ))) j=0 j=0 µ( (A i \ B i )) + µ( (B j \ A j )) j=0 µ(a i \ B i ) + µ(b j \ A j ) j=0 = 0, so A i µ B j. Thus [ A i ] µ = [ B j ] µ and is well-defined. By the above, j=0 [A c n] µ = [Bn] c µ for all n N, so µ(( A c i) ( Bj)) c = 0. Then µ(( A i ) ( B j )) = j=0 20 j=0 j=0

26 µ(( A c i) c ( Bj) c c ) = µ(( A c i) ( Bj)) c = 0, so A i µ j=0 [ B j ] µ and is well-defined. j=0 j=0 B j. Thus [ A i ] µ = j= Proposition. The set ˆB with 0 = [ ]µ, 1 = [X] µ, and operations,, and c is a Boolean σ-algebra. Proof. Let [A] µ, [B] µ, and [C] µ ˆB. (1) [A] µ [A] µ = [A A] µ = [A] µ (2) [A] µ [A] µ = [A A] µ = [A] µ (3) [A] µ [B] µ = [A B] µ = [B A] µ = [B] µ [A] µ (4) [A] µ [B] µ = [A B] µ = [B A] µ = [B] µ [A] µ (5) ([A] µ [B] µ ) [B] µ = [A B] µ [B] µ = [(A B) B] µ = [B] µ (6) ([A] µ [B] µ ) [B] µ = [A B] µ [B] µ = [(A B) B] µ = [B] µ (7) ([A] µ [B] µ ) [C] µ = [A B] µ [C] µ = [(A B) C] µ = [A (B C)] µ = [A] µ [(B C)] µ = [A] µ ([B] µ [C] µ ) (8) ([A] µ [B] µ ) [C] µ = [A B] µ [C] µ = [(A B) C] µ = [A (B C)] µ = [A] µ [B C] µ = [A] µ ([B] µ [C] µ ) (9) [A] µ ([B] µ [C] µ ) = [A] µ [B C] µ = [A (B C)] µ = [(A B) (A C)] µ = [A B] µ [A C] µ = ([A] µ [B] µ ) ([A] µ [C] µ ) (10) [A] µ ([B] µ [C] µ ) = [A] µ [B C] µ = [A (B C)] µ = [(A B) (A C)] µ = [A B] µ [A C] µ = ([A] µ [B] µ ) ([A] µ [C] µ ) (11) [A] µ [A] c µ = [A] µ [A c ] µ = [A A c ] µ = [X] µ (12) [A] c µ [A] µ = [A c A] µ = [ ] µ (13) [ ] µ [X] µ since µ( X) = 1 > 0 21

27 Let [A i ] µ ˆB N. Let B = [A i ] µ. For all n N, [A n ] µ B = [A n ] µ ( [A i ] µ ) = [A n ] µ [ A i ] µ = [ (A n A i )] µ = [(A n A n ) ( (A n A i ))] µ = [A n ] µ, so [A n ] µ B. Thus B is an upper bound for [A i ] µ. Now suppose [C] µ = Ĉ ˆB is another upper bound for [A i ] µ. Then B Ĉ = ( [A i ] µ ) [C] µ = [ A i ] µ [C] µ = [( A i ) C] µ = [ (A i C)] µ = [(A i C)] µ = ([A i ] µ [C] µ ) = ([A i ] µ Ĉ). Since for all i N, [A i ] µ Ĉ, we have B Ĉ = [A i ] µ = B, so B Ĉ. Hence B is the least upper bound of [A i ], so ˆB is a Boolean σ-algebra. i n Let ˆµ : ˆB [0, 1] such that ˆµ([A] µ ) = µ(a) Proposition. ˆµ is well-defined and strictly positive. Proof. Let A, C B such that [A] µ = [C] µ. By definition, A µ C, so µ(a \ C) + µ(c \ A) = µ(a C) = 0. Then µ(a \ C) = 0 = µ(c \ A), so µ(a) = µ(a) + µ(c \ A) = µ(a (C \ A)) = µ(a C) = µ(c (A \ C)) = µ(c) + µ(a \ C) = µ(c). Hence ˆµ([A] µ ) = µ(a) = µ(c) = ˆµ([C] µ ), so ˆµ is well-defined. Suppose ˆµ([A] µ ) = 0. Then µ(a) = 0, so µ(a ) = 0. Thus A, so [A] µ = [ ] µ = Proposition. ˆµ is countably additive. Proof. Let (B n ) n N ˆB N be pairwise disjoint, i.e. for all m n, B m B n = 0. For each n N, choose A n B such that [A n ] µ = B n. Let m, n N. Then [ ] µ = 0 = [A m ] µ [A n ] µ = [A m A n ] µ, so A m A n. Therefore, 0 = µ((a m A n ) ) = µ(((a m A n ) \ ) ( \ (A m A n ))) µ(((a m A n ) \ )) + (( \ (A m A n ))) = µ(a m A n ) + 0, so µ(a m A n ) = 0. To use the countable additivity of µ, we will construct a disjoint sequence (D n ) n N in B such that µ(d n ) = µ(a n ) for all n N. Let D 0 = A 0 and D n = A n \ ( n 1 22 A i ) for all

28 n > 0 and let n N. Then n 1 µ(d n ) = µ(a n \ ( A i )) n 1 = µ(a n \ (A n ( A i ))) n 1 = µ(a n \ ( (A n A i ))) n 1 = µ(a n ) µ( (A n A i )) since n 1 µ(a n ) µ(a n A i ) n 1 (A n A i ) A n = µ(a n ) since µ(a n A i ) = 0 for all i N, so ˆµB i = ˆµ[A i ] µ = = µa i µd i = µ( D i ) by countable additivity of µ = µ( A i ) = ˆµ([ A i ] µ ) = ˆµ( i N[A i ] µ ) by definition 23

29 = ˆµ( i N B i ). Thus ˆµ is countably additive. Since ˆµ([ ] µ ) = 0, ˆµ([X] µ ) = µ(x) = 1, and ˆµ is countably additive, ˆµ is a probability measure. Let ˆd : ˆB R such that ˆd([A] µ, [B] µ ) = ˆµ([A] µ [B] µ ) Proposition. ˆd is a metric on ˆB. Proof. Let [A] µ, [B] µ, [C] µ ˆB. (1) non-negativity: ˆd([A] µ, [B] µ ) = ˆµ([A] µ [B] µ ) 0 (2) coincidence: ˆd([A]µ, [B] µ ) = 0 ˆµ([A] µ [B] µ ) = 0 ˆµ([A B] µ ) = 0 µ(a B) = 0 A µ B [A] µ = [B] µ (3) symmetry: ˆd([A] µ, [B] µ ) = ˆµ([A] µ [B] µ ) = ˆµ([A B] µ ) = µ(a B) = µ(b A) = ˆµ([B A] µ ) = ˆµ([B] µ [A] µ ) = ˆd([B] µ, [A] µ ) (4) triangle inequality: ˆd([A] µ, [B] µ ) + ˆd([B] µ, [C] µ ) = ˆµ([A] µ [B] µ ) + ˆµ([B] µ [C] µ ) = µ(a B) + µ(b C) = µ(a \ B) + µ(b \ A) + µ(b \ C) + µ(c \ B) since (A \ B) (B \ A) = = (B \ C) (C \ B) µ(a \ C) + µ(c \ A) since A \ C (A \ B) (B \ C) and C \ A (C \ B) (B \ A) = µ(a C) since (A \ C) (C \ A) = 24

30 = ˆµ([A] µ [C] µ ) = ˆd([A] µ, [C] µ ) Proposition. ( ˆB, ˆd) is a complete metric space. Proof. (323F in [9]) Let a n ˆB N be Cauchy with respect to ˆd. For all k N there exists a minimal N k N such that for all m, n N k, ˆd(a m, a n ) < 2 k. Note that N k k N is a nondecreasing sequence. For each k, let c k = a Nk. Then ˆd(c k=0 k, c k+1 ) k=0 2 k <. Let p 0 = n N m n c m and let p 1 = n N m n c m. Note that since ˆB is σ-complete, p 0, p 1 ˆB. First, I will show that p 0 = p 1. Let α n = ˆµ(c n c n+1 ) and β n = α k for all n N. Since ˆd(c n, c n+1 ) = ˆµ(c n c n+1 ) = α n for all n, n=0 α n = k=0 ˆd(c k, c k+1 ) <, so lim n β n = 0. Let b n = n k c k c k+1. Then for all n, by countable additivity of ˆµ. ˆµ(b n ) = ˆµ( m n c m c m+1 ) k=n ˆµ(c m c m+1 ) = β n m=n Fix n N and let n m. I will show that c m c n c m c n = 0 = c k c k+1 = c k c k+1. For m n + 2, k=n = ( ( n k<m 2 n k<m 2 n k<m n k<m n k<m c k c k+1. If m = n, then c k c k+1. If m = n + 1, then c m c n = c n+1 c n = c k c k+1 c k c k+1 ) ((c m 2 c m 1 ) (c m 1 c m )) c k c k+1 ) (c m 2 c m ) by

31 = ( c k c k+1 ) ((c m 3 c m 2 ) (c m 2 c m )) n k<m 3 ( c k c k+1 ) (c m 3 c m ) by n k<m 3 c n c m. Thus, c m c n c k c k+1 n k c k c k+1 = b n. Since c m c n b n, we have n k<m (c n \b n )\c m = (c n b c n) c c m = (c n c c m) b c n = (c n \c m )\b n (c n c m )\b n b n \b n = 0, so (c n \ b n ) \ c m = 0. By 4.1.6, c n \ b n c m. We can also use c m c n b n to show that c m c n b n. Since c m \ c n b n, (c m \ c n ) b n = c m \ c n, so ((c m \ c n ) b n ) c n = (c m \ c n ) c n. Expanding both sides, we get (c m c c n b n ) c n = (c m c c n) c n, so (c m c n ) (b n c n ) = (c m c n ) 1 (b n c n ) = (c m c n ) (c c n c n ) (b n c n ) = (c m c n ) 1 = c m c n. Thus c m c m c n c n b n. Putting this together, we get c n \ b n c m c n b n. Since this holds for every m n, c n \ b n c m c m c n b n m k m k for all k n. Since this holds for every k n, c n \b n k N m k c m k N m k c m c n b n. In other words, since p 0 = k N m k c m and p 1 = k N m k c m, we have c n \ b n p 0 p 1 c n b n. I will use this to show that c n p i b n for i = 0, 1. Let i {0, 1}. Since c n \ b n p i, we have (c n \ p i ) \ b n = (c n \ b n ) \ p i p i \ p i = 0, so (c n \ p i ) \ b n = 0. By 4.1.6, 26

32 c n \ p i b n. Since p i c n b n, p i (c n b n ) = p i, so (p i \ c n ) \ b n = p i c c n b c n = p i (c n b n ) c = (p i (c n b n )) (c n b n ) c = 0. By 4.1.6, p i \ c n b n. Thus c n p i = (c n \ p i ) (p i \ c n ) b n. Another application of gives us p 1 \ p 0 b n : (p 1 \ p 0 ) \ b n = p 1 p c 0 b c n = (p 1 (c n b n )) p c 0 b c n since p 1 c n b n = p 1 p c 0 (b c n (c n b n )) = p 1 p c 0 ((b c n c n ) (b c n b n )) = p 1 p c 0 ((b c n c n ) 0) = p 1 p c 0 (c n \ b n ) = p 1 p c 0 ((c n \ b n ) p 0 ) since c n \ b n p 0 = p 1 (c n \ b n ) (p c 0 p 0 ) = p 1 (c n \ b n ) 0 = 0. Since c n p i b n holds for every n, lim n ˆµ(c n p i ) lim n ˆµ(b n ) lim n β n = 0 for i = 0, 1. Furthermore, since p 0 p 1, ˆµ(p 1 p 0 ) = ˆµ(p 1 \ p 0 ) lim n ˆµ(b n ) = 0. In fact, this implies that p 0 = p 1. To see this, let P 0, P 1 B such that p 0 = [P 0 ] µ, p 1 = [P 1 ] µ. Then 0 = ˆµ(p 1 p 0 ) = ˆµ([P 1 ] µ [P 0 ] µ ) = ˆµ([P 1 P 0 ] µ ) = µ(p 1 P 0 ), so p 0 = [P 0 ] µ = [P 1 ] µ = p 1. 27

33 Now let p = p 0 = p 1. I will show that lim n a n = lim n c n = p. Since lim n ˆd(cn, p) = lim n ˆµ(c n p) = 0, c n n N converges to p with respect to ˆd. Let ɛ > 0. Since c n n N converges to p, there exists M 1 N such that for all k > M 1, ˆd(ck, p) < ɛ/2. Since a n n N is Cauchy, there exists M 2 N such that for all m, n > M 2, ˆd(am, a n ) < ɛ/2. Let M = max(m 1, M 2 ). Then k > M implies ˆd(a k, p) ˆd(a k, a Nk ) + ˆd(a Nk, p) = ˆd(a k, a Nk ) + ˆd(c k, p) < ɛ/2 + ɛ/2 = ɛ, so the original sequence a n n N converges to p ˆB. Since a n n N was an arbitrary Cauchy sequence, ( ˆB, ˆd) is a complete metric space. Since ( ˆB, ˆd) is a complete metric space, M = ( ˆB, 0, 1, c,,, ˆµ) as constructed above is a metric structure. We call such an M a probability algebra, or, in the context of its being a metric structure, a probability structure Here are some examples of probability structures. (1) Let (X, B, µ) be a probability space where X = {a 1, a 2,..., a 10 }, B = P(X), and µ(a n ) = 1/10 for all n = 1,..., 10. Since there are no nontrivial events of measure zero, µ = ˆµ. The distance between two events is the probability that exactly one of them happens. (2) Let (X, B, µ) be a probability space where X is the unit interval [0, 1], B is the σ-algebra of Borel sets, and µ is the Lebesgue measure on [0, 1]. ˆB identifies events whose symmetric difference is a null set. (3) Let X = {0, 1} N, the set of infinite sequences of 0s and 1s. B is the σ-algebra generated by sets of the form {(x 1, x 2,...) X : (x 1,..., x n ) = (a 1,..., a n )}. If we think of X as the set of countably many tosses of a fair coin, then a generator 28

34 of B is the event that the first n tosses are a fixed sequence (a 1,..., a n ). Then µ({(x 1, x 2,...) X : (x 1,..., x n ) = (a 1,..., a n )}) = 2 n. Two distinct generator sets have symmetric difference > 0, and thus remain distinct in ˆB. Definition. The signature associated with probability structures is L P r = (R, F, η, G), where (1) R = {ˆµ, ˆd} (2) F = {0, 1, c,, } (3) 0 = η(0) = η(1), 1 = η(ˆµ) = η( c), 2 = η( ˆd) = η( ) = η( ) (4) G = {δ c, δˆµ, δ, δ, δ ˆd} where δ c(ɛ) = δˆµ (ɛ) = ɛ and δ (ɛ) = δ (ɛ) = δ ˆd(ɛ) = ɛ/ Proposition. As defined above, δ c, δˆµ, δ, δ, δ ˆd are indeed moduli of uniform continuity. Proof. δ c: Let ɛ (0, 1] and [A] µ, [B] µ ˆB. Suppose d([a] µ, [B] µ ) < δ c(ɛ) = ɛ. Then d([a] c µ, [B] c µ) = d([a c ] µ, [B c ] µ ) = µ(a c B c ) = µ(a B) = d([a] µ, [B] µ ) < ɛ. δˆµ : Let ɛ (0, 1] and [A] µ, [B] µ ˆB. Suppose d([a] µ, [B] µ ) < δˆµ (ɛ) = ɛ. Since 2ˆµ(a b) 2ˆµ(a), we have ˆµ(a) + 2ˆµ(a b) ˆµ(a), so ˆµ(a b) = ˆµ(a) ˆµ(b) + 2ˆµ(a b) ˆµ(a) ˆµ(b). Since 2ˆµ(a b) 2ˆµ(b), we have ˆµ(b) ˆµ(b) 2ˆµ(a b), so ˆµ(a) ˆµ(b) ˆµ(a) + ˆµ(b) 2ˆµ(a b) = ˆµ(a b). Putting this together, we get d(ˆµ(a), ˆµ(b)) = ˆµ(a) ˆµ(b) ˆµ(a b) < ɛ. δ : Let ɛ (0, 1] and ([A 1 ] µ, [A 2 ] µ ), ([B 1 ] µ, [B 2 ] µ ) ˆB ˆB. Suppose dmax(([a 1 ] µ, [A 2 ] µ ), ([B 1 ] µ, [B 2 ] µ )) < δ (ɛ) where δ (ɛ) = ɛ/2. Then for i = 1, 2, ɛ/2 > d([a i ] µ, [B i ] µ ) = µ(a i B i ) = µ(a i \ B i ) + µ(b i \ A i ). Since (A 1 A 2 ) \ (B 1 B 2 ) = (A 1 \ (B 1 B 2 )) (A 2 \ (B 1 B 2 )) 29

35 (A 1 \ B 1 ) (A 2 \ B 2 ) and (B 1 B 2 ) \ (A 1 A 2 ) = (B 1 \ (A 1 A 2 )) (B 2 \ (A 1 A 2 )) (B 1 \ A 1 ) (B 2 \ A 2 ), we have d([a 1 ] µ [A 2 ] µ, [B 1 ] µ [B 2 ] µ ) = d([a 1 A 2 ] µ, [B 1 B 2 ] µ ) = µ((a 1 A 2 ) (B 1 B 2 )) = µ((a 1 A 2 ) \ (B 1 B 2 )) + µ((b 1 B 2 ) \ (A 1 A 2 )) µ((a 1 \ B 1 ) (A 2 \ B 2 )) + µ((b 1 \ A 1 ) (B 2 \ A 2 )) µ(a 1 \ B 1 ) + µ((a 2 \ B 2 ) + µ(b 1 \ A 1 ) + µ(b 2 \ A 2 ) = (µ(a 1 \ B 1 ) + µ(b 1 \ A 1 )) + (µ((a 2 \ B 2 ) + µ(b 2 \ A 2 )) = µ(a 1 B 1 ) + µ(a 2 B 2 ) = d([a 1 ] µ, [B 1 ] µ ) + d([a 2 ] µ, [B 2 ] µ ) 2max{d([A 1 ] µ, [B 1 ] µ ), d([a 2 ] µ, [B 2 ] µ )} < 2(ɛ/2) = ɛ. δ : Let ɛ (0, 1] and ([A 1 ] µ, [A 2 ] µ ), ([B 1 ] µ, [B 2 ] µ ) ˆB ˆB. Suppose dmax(([a 1 ] µ, [A 2 ] µ ), ([B 1 ] µ, [B 2 ] µ )) < δ (ɛ) where δ (ɛ) = ɛ/2. Since (A 1 A 2 ) \ (B 1 B 2 ) = ((A 1 A 2 ) \ B 1 ) ((A 1 A 2 ) \ B 2 ) (A 1 \ B 1 ) (A 2 \ B 2 ) and (B 1 B 2 ) \ (A 1 A 2 ) = ((B 1 B 2 ) \ A 1 ) ((B 1 B 2 ) \ A 2 ) (B 1 \ A 1 ) (B 2 \ A 2 ), we have d([a 1 ] µ [A 2 ] µ, [B 1 ] µ [B 2 ] µ ) = d([a 1 A 2 ] µ, [B 1 B 2 ] µ ) 30

36 = µ((a 1 A 2 ) (B 1 B 2 )) = µ((a 1 A 2 ) \ (B 1 B 2 )) + µ((b 1 B 2 ) \ (A 1 A 2 )) µ((a 1 \ B 1 ) (A 2 \ B 2 )) + µ((b 1 \ A 1 ) (B 2 \ A 2 )) µ(a 1 \ B 1 ) + µ(a 2 \ B 2 ) + µ(b 1 \ A 1 ) + µ(b 2 \ A 2 ) = (µ(a 1 \ B 1 ) + µ(b 1 \ A 1 )) + (µ((a 2 \ B 2 ) + µ(b 2 \ A 2 )) = µ(a 1 B 1 ) + µ(a 2 B 2 ) = d([a 1 ] µ, [B 1 ] µ ) + d([a 2 ] µ, [B 2 ] µ ) 2max{d([A 1 ] µ, [B 1 ] µ ), d([a 2 ] µ, [B 2 ] µ )} < 2(ɛ/2) = ɛ. δ ˆd: Let ɛ (0, 1] and ([A 1 ] µ, [A 2 ] µ ), ([B 1 ] µ, [B 2 ] µ ) ˆB ˆB. Suppose dmax(([a 1 ] µ, [A 2 ] µ ), ([B 1 ] µ, [B 2 ] µ )) < δ ˆd(ɛ) where δ ˆd(ɛ) = ɛ/2. By the triangle inequality, ˆd(([A 1 ] µ, [A 2 ] µ )) ˆd(([B 1 ] µ, [B 2 ] µ )) ˆd(([A 1 ] µ, [B 1 ] µ )) + ˆd(([B 1 ] µ, [B 2 ] µ )) + ˆd(([B 2 ] µ, [A 2 ] µ )) ˆd(([B 1 ] µ, [B 2 ] µ )) = ˆd(([A 1 ] µ, [B 1 ] µ )) + ˆd(([B 2 ] µ, [A 2 ] µ )) < ɛ/2 + ɛ/2 = ɛ. 31

37 since max{d([a 1 ] µ, [B 1 ] µ ), d([a 2 ] µ, [B 2 ] µ )} < ɛ/2. Similarly, ˆd(([A 1 ] µ, [A 2 ] µ )) ˆd(([B 1 ] µ, [B 2 ] µ )) ˆd(([A 1 ] µ, [A 2 ] µ )) ( ˆd(([B 1 ] µ, [A 1 ] µ )) + ˆd(([A 1 ] µ, [A 2 ] µ )) + ˆd(([A 2 ] µ, [B 2 ] µ ))) = ( ˆd(([B 1 ] µ, [A 1 ] µ )) + ˆd(([A 2 ] µ, [B 2 ] µ ))) > (ɛ/2 + ɛ/2) = ɛ. Putting this together, we get d( ˆd(([A 1 ] µ, [A 2 ] µ )), ˆd(([B 1 ] µ, [B 2 ] µ ))) = ˆd(([A 1 ] µ, [A 2 ] µ )) ˆd(([B 1 ] µ, [B 2 ] µ )) < ɛ. Notation. A Boolean ring is a ring (A, +, ) where a 2 = a for all a A. We can represent any Boolean algebra A = (A, 0, 1,,, c) as a Boolean ring (A, +, ) by taking a + b := a b and ab := a b. Conversely, given a Boolean ring (A, +, ), we can view it as a Boolean algebra by taking a b := a + b + ab, a b := ab, and a c := 1 + a. Depending on the context, one representation may be preferred over the other. Definition. (311F in [9]) Given a Boolean ring B, the Stone space of B is the set Z of nonzero ring homomorphisms from B to Z 2, i.e. z Z a, b B z(a + b) = z(a) + z(b), z(ab) = z(a)z(b), z(1) = 1 Z2. For each a B, let â = {z Z : z(a) = 1}. The Stone representation of B is the canonical map a â : B P(Z) Lemma. (a special case of 311D in [9]) Let B be a Boolean ring and let a B be nonzero. Then there exists z Z such that z(a) = 1. Proof. Let I = {J : J is an ideal of B and a / J}. First I will use Zorn s lemma on (I, ) to show that I has a maximal element. Note that I is nonempty because {0} I. Let J 32

38 be a nonempty totally ordered subset of I and let J 0 = J. Let m, n J 0. Then there exist J 1, J 2 J such that m J 1 and n J 2. Since J is totally ordered, J 1 J 2 or J 2 J 1. Without loss of generality, suppose J 1 J 2, so m, n J 2. Let r B. Since J 2 is an ideal, 0, m + n, rm J 2 J 0, so J 0 is an ideal. Since a / J 0, J 0 I, so J 0 is an upper bound for J. Since J was arbitrary, by Zorn s lemma, I has a maximal element, which we will denote by K. Next, I will show that for all m, n B \ K, m + n K. For all b B, let K b = {c B : bc K}. Let b B. Since K is an ideal, for all n K, we have nb K, so n K b. Thus K K b. If n, n K b and c B, then b(n + n ) = bn + bn K and b(nc) K, so n + n, nc K b. Since 0 K K b, K b is an ideal. If a / K b, then K b I. Since K is maximal in I and K K b, we have K = K b. In particular, a / K a, since a 2 = a / K, so K a = K. Now let b, c B \ K. Since b / K = K a, we have ba = ab / K, so a / K b. Hence K b = K. Since c / K = K b, we have bc / K. Now bc(b + c) = b 2 c + bc 2 = bc + bc = 0 K, so b + c K bc. Since bc / K and for all e / K, K = K e, we have K = K bc, so b + c K. Now I will construct a ring homomorphism z : B Z 2 such that z(a) = 1. For all d B, let 0 if d K z(d) := 1 if d B \ K. Let b, c B. If b, c K, then b + c, bc K, so z(b + c) = 0 = = z(b) + z(c) and z(bc) = 0 = 0 0 = z(b)z(c). If b K and c B \ K, then b + c / K, since otherwise c = b + (b + c) K, so z(b + c) = 1 = = z(b) + z(c). Since bc K, z(bc) = 0 = 0 1 = z(b)z(c). If b B \ K and c K, then z(b + c) = 1 = = z(b) + z(c) 33

39 and bc K, z(bc) = 0 = 1 0 = z(b)z(c). Finally, if b, c B \ K, then as shown in the previous paragraph, b + c K and bc / K, so z(b + c) = 0 = = z(b) + z(c) and z(bc) = 1 = 1 1 = z(b)z(c). Hence z is a ring homomorphism and, since a / K, z(a) = Proposition. (311E in [9]) The Stone representation of a Boolean ring B is an injective ring homomorphism a â : (B, +, ) (P(Z),, ). Proof. Let B be a Boolean algebra and let a, b B. Then â + b = {z : z(a + b) = 1} = {z : z(a) + z(b) = 1} = {z : z(a) = 0 and z(b) = 1, or z(a) = 1 and z(b) = 0} = (â (ˆb) c ) (ˆb (â) c ) = â ˆb and âb = {z : z(ab) = 1} = {z : z(a)z(b) = 1} = {z : z(a) = 1 and z(b) = 1} = â ˆb, so a â is a ring homomorphism. To see that this ring homomorphism is injective, let a B be nonzero. By 4.10, there exists z Z such that z(a) = 1. Thus z â, so â. Hence the kernel of this homomorphism is {0}, so a â is injective. By 4.11, any Boolean ring B is isomorphic to its image {â : a B} under the Stone representation. In particular, ˆ commutes with Boolean operations: for all a, b B, (1) â b = a b (a b) = a + b + ab = â b + ab = â ˆb âb = â ˆb (2) â b = âb = â ˆb (3) â \ b = â bc = a(1 + b) = â + ab = â âb = â (â ˆb) = (â \ (â ˆb)) ((â ˆb) \ â) = â \ (â ˆb) = â \ ˆb (4) â b = â + b = â ˆb. 34

40 Also note that a â : B P(Z) respects the order relation on B, i.e. for all a, b B, a b â ˆb. Finally, we have ˆ1 = Z: Let z Z. By definition, z a is nonzero ring homomorphism, so there exists a B such that z(a) = 1. Then 1 = z(a) = z(a 1) = z(a1) = z(a)z(1) = 1z(1) = z(1), so z ˆ1 = Z Lemma. Let B be a Boolean algebra and let Z be its Stone space. Let T be {G Z : z G a B z â G}. Then T is a topology on Z under which Z is a Hausdorff space. Proof. First, we show that T is a topology. Let z Z. Since z 0, there exists a B such that z(a) = 1, so z â. Thus Z E. Now let â, ˆb E and z â ˆb. By 311G, â ˆb = â b E, so in particular, z â b â ˆb. Thus E is a base for some topology on Z. Since T is defined to be the union of elements of E, T is the topology generated by E. To see that (Z, T ) is Hausdorff, let w, z Z be distinct. Then there exists a B such that z(a) w(a), say z(a) = 1 and w(a) = 0. Since w is nonzero, there exists b B such that w(b) = 1. Then w(b \ a) = w(b a c ) = w(b(1 + a)) = w(b + ba) = w(b) + w(b)w(a) = 1, so w b \ a T. Since z(a) = 1, z â T. Also, ( b \ a) â = (b \ a) a = b (ac a) = ˆ0 = 1 1 = ˆ1 ˆ1 =, so b \ a and â are disjoint neighborhoods of w and z, respectively. Hence (Z, T ) is Hausdorff (311I in [9]) The Stone space Z of a Boolean algebra B is compact. E = {â : a B} is precisely the set of compact and open subsets of Z. Definition. Let X be a set and let B be a σ-algebra of subsets of X. A B is a σ-ideal if it satisfies the following properties: 35

41 (1) A (2) For all A A, B B, if B A, then B A (3) If A n n N A N, then n N A n A Definition. Let X be a topological space. For A X, denote the closure of A by A and the interior of A by inta. A X is nowhere dense if inta =. A X is meager if there exist countably many B n X such that A = n N B n and each B n is nowhere dense. Definition. (312F in [9]) Given Boolean algebras A and B, a function π : A B is a Boolean homomorphism if π(a b) = π(a) π(b) and π(a b) = π(a) π(b) for all a, b A, and π(1 A ) = 1 B Lemma. (313Ca in [9]) Let A be a σ-complete Boolean algebra and let Z be its Stone space. Let a n be a sequence in A and let a = sup n N a n A. Then n N ân = â. Proof. For all n, a n a, so â n â by Then â n N ân = n N ân, since â is closed. To show that â n N ân, suppose for the sake of contradiction that â\ n N ân. Since n N ân is closed, â \ n N ân = â ( n N ân) c is open, so there exists a nonzero base element ˆb â \ n N ân. Then for all n, â n â \ ˆb = â \ b, so by 4.11, a n a \ b, which makes a \ b an upper bound for a n. Since b < a = sup n N a n, we have a contradiction. Thus n N ân = â. Let X be a topological space. Recall that X is a Baire space if the intersection of every countable collection of dense open sets in X is dense, and that X is a Hausdorff space if distinct points in X have disjoint open neighborhoods. In the following lemmas, we will use a weakened version of the Baire category theorem: If X is a compact Hausdorff space, then X is a Baire space. 36

42 4.15. Lemma. In a compact Hausdorff space, if a set is meager and open, then it is empty. Proof. Let X be a compact Hausdorff space. By the Baire category theorem, X is a Baire space. Let E H be open and meager. Since E is meager, we can write E as the union of countably many nowhere dense sets {M n } n N. Fix n N. By definition, int(m n ) =, so X = X \ int(m n ) = X \ M n = int(x \ M n ). Thus int(x \ M n ) is dense in X. Also int(x \ M n ) is open, so since X is a Baire space, int(x \ M n ) is dense. Then int(x \ M n ) (X \ M n ) = X \ M n = X \ E. Since E is open, X \ E is n N closed, so X = n N n N n N int(x \ M n ) X \ E = X \ E. Hence E is empty Lemma. In any topological space Z, the family M of meager subsets of Z forms a σ-ideal. n N Proof. Let Z be a topological space and let M be the family of meager subsets of X. Since int =, M. Let A M and B A. Let A n n N be a sequence of subsets of Z such that n N A n = A and for all n, inta n =. Then for all n, intb A n inta n =. Since n N B A n = B n N A n = B A = B, B M. Now let A n n N M N. For each n, there exists A i n i N a sequence of subsets of Z such that i N Ai n = A n and for all i, inta i n =. Since A i n n,i N is countable, we can re-index it to a sequence B k k N. Then for all k, B k is nowhere dense, so n N A n = i,n N Ai n = k N B k M. The following theorem is the Loomis-Sikorski representation of a σ-complete Boolean algebra. It will be used in the proof of the Stone representation theorem of a measure algebra in Theorem. (314M in [9]) Let A be a σ-complete Boolean algebra, and Z its Stone space. Let E = {â : a A}, and let M be the σ-ideal of meager subsets of Z. Then 37

43 Σ = {E A : E E, A M} is a σ-algebra of subsets of Z, M is a σ-ideal of Σ, and A is isomorphic, as Boolean algebra, to Σ/M. Proof. First, I will show that Σ is a σ-algebra and M is a σ-ideal of Σ. Since = ˆ0 E, Σ. Let F Σ. There exist a B, A M such that F = â A. Since (â) c = â c E, F c = (â A) c = ((â A c ) (â c A)) c = (â A c ) c (â c A) c = (â c A) (â A c ) = ((â c A) A c ) ((â c A) â) = [(â c A c ) (A A c )] [(â c â) (A â)] = (â c A c ) (A â) = (â c \ A) (A \ â c ) = â c A Σ. Now let F n n N Σ N. For each n, let a n B, A n M such that F n = â n A n. Since B is σ-complete, a := sup n N a n B. By 4.14, n N ân = â. Let C := â\ n N ân. I will show that C has empty interior and is nowhere dense. Since each â n is open, n N ân is open, so C = â ( n N ân) c is closed. Since n N ân is open, n N ân = int( n N ân), so n N ân = cl( n N ân) \ int( n N ân) = cl( n N ân) \ n N ân = C. Suppose C contains an open set U. Then U and n N ân are disjoint. Since 38

44 U is open and in the boundary of n N ân, U and n N ân has nonempty intersection, contradiction. Hence int(c) =. Since C is closed, C = C, so int(c) =, i.e. C is nowhere dense. Now let A = â n N F n. I will show that A C ( n N A n). Let x A. First suppose x ( n N F n) \ â. Then there exists n N such that x F n = â n A n. In other words, x â n A n and x / â n A n. Since â n â and x / â, we must have x A n C ( n N A n). Next, suppose x â \ n N F n. I will show that if x / n N A n, then x C, so suppose x / n N A n. Since x / n N F n = n N ân A n, for all n either x â n A n or x / â n A n. By assumption, x / â n A n, so x / â n A n. Thus x / n N ân, so x â \ n N ân = C. Thus A C ( n N A n). Since M is a σ-ideal, n N A n M, so since C is nowhere dense, C ( n N A n) M. Finally, since A C ( n N A n), A M. Thus, n N F n = n N F n = (â â) n N F n = â A Σ, so Σ is closed under countable unions. Hence, Σ is a σ-algebra. Since = ˆ0 E, M = { A : A M} Σ, so by 4.16 M is a σ-ideal of Σ. Next, let be a relation on Σ such that for all F 1, F 2 Σ, F 1 F 2 K M such that F 1 = F 2 K. I will show that is an equivalence relation. Let F 1, F 2, F 3 Σ. Since F 1 = F 1 and M, is reflexive. Suppose F 1 = F 2 K for some K M. Then F 1 K = (F 2 K) K = F 2 = F 2, so is symmetric. Finally, let J, K M such that F 1 = F 2 K and F 2 = F 3 J. Then F 1 = F 2 K = (F 3 J) K = F 3 (J K) and since J K M, is transitive. Denote the equivalence class of F Σ under by [F ] M and let Σ/M = {[F ] M : F Σ}. Define the operations M and M on Σ/M as follows: for all f 1 = [F 1 ] M, f 2 = 39